如何通过Windows缩放处理SendInput鼠标移动？ [复制]

Question

此问题已经在这里有了答案：

• How to configure an app to run correctly on a machine with a high DPI setting (e.g. 150%)?4个答案

所以我找不到任何可以帮助我解决此问题的东西，这使我发疯了。

使用Windows Input Simulator (C# SendInput Wrapper - Simulate Keyboard and Mouse)

当我的软件用户碰巧遇到非标准的100％窗口缩放InputSimulator.Mouse.MoveMouseTo(toX, toY);时，将移至错误的位置。

我目前的计算方式：

int toX = (int)((65535.0f * (newX / (float)screen.Bounds.Width)) + 0.5f);
int toY = (int)((65535.0f * (newY / (float)screen.Bounds.Height)) + 0.5f);

例如，如果用户将缩放比例设置为125％，我将如何计算正确的100％缩放位置？

提前感谢。

Answer 1

答案很简单，只是一个令人沮丧的问题要解决。

首先检测用户是否已像这样应用缩放：

[DllImport("gdi32.dll")]
public static extern int GetDeviceCaps(IntPtr hdc, int nIndex);

public enum DeviceCap { VERTRES = 10, DESKTOPVERTRES = 117 }
public float GetScalingFactor()
{
    Graphics g = Graphics.FromHwnd(IntPtr.Zero);
    IntPtr desktop = g.GetHdc();
    int LogicalScreenHeight = GetDeviceCaps(desktop, (int)DeviceCap.VERTRES);
    int PhysicalScreenHeight = GetDeviceCaps(desktop, (int)DeviceCap.DESKTOPVERTRES);

    float ScreenScalingFactor = (float)PhysicalScreenHeight / (float)LogicalScreenHeight;

    return ScreenScalingFactor; // 1.25 = 125%
}

然后存储比例尺并将其除以初始坐标。

ScreenScale = Utils.GetScalingFactor();
int toX = (int)((65535.0f * ((x / ScreenScale) / (float)wowScreen.Bounds.Width)) + 0.5f);
int toY = (int)((65535.0f * ((y / ScreenScale) / (float)wowScreen.Bounds.Height)) + 0.5f);